Where is the Center of SHM in a Light Elastic String System?

[shalikadm]

Homework Statement

One end of a light elastic string having a natural length of 'a' and elastic modulus 'mg' is attached to 'm' mass and the other end to point 'O'. Then the mass is projected upwards.Where is the center of SHM?

Homework Equations

ω=angular velocity
F=ma
from Hooke's law,
T=λx/l
T=tension of the string
λ=elastic modulus
l=natural length
x=extention

The Attempt at a Solution

There's no any equilibrium in the motion above point 'O' as T and mg are towards the same direction.But after point 'A' I think that there's a SHM as the string is extended.In the calculation, it says that center of SHM is at point 'O'..How can it be true ?...No equilibrium there !...It must be somewhere under that point 'O' where the two forces on the mass are opposing each other..Or where have I got it wrong ?
thanks !

 

[DryRun]

I think this question has been posted in the wrong board - it belongs in the Physics section.

 

[HallsofIvy]

No, as stated this is a math problem. Taking x= 0 at point O, where x is the position of the weight, positive downward, the total force on the weight mg and $-\lambda(x- a)$. The position at time t is given by $d^2x/dt^2= mg- \lambda(x- a)$. If $\lambda$ happens to be equal to mg, this is $d^2x/dt^2= mg((1+a)- x)$ or $d^2x/dt^2+ mgx= (1+a)mg$.

The general solution to that equation is
$$x(t)= Ccos(\sqrt{mg}t)+ Dsin(\sqrt{mg}t)+ 1+ a$$

You can find the max and min of that and then the center of motion will be 1/2 of those.

 

[shalikadm]

I think that there are some mistakes in your calculation...

the total force on the weight mg and −λ(x−a)

It must be mg and -λ(x-a)/a...[see also that λ's dimensions is kgms^-2..so (x-a)/a must have no dimensions]

The position at time t is given by $d^2x/dt^2= mg- \lambda(x- a)$

$d^2x/dt^2$ is an acceleration...So the right side of the equation it must be
$d^2x/dt^2=g- \lambda(x- a)/a$(as the dimensions of g and λ(x-a)/a must be ms^-2)
and so..$d^2x/dt^2= mg(2a- x)$
and so the general solution must be something like this...
$$x(t)= Ccos(\sqrt{g/a}t)+ Dsin(\sqrt{g/a}t)+ 2a$$
Any way
we don't need to find general solution...we can find the center of motion from $d^2x/dt^2=o$..so its $x=2a$
You have considered the downside of the motion..I want to consider the upside motion..Is there an SHM in the upside motion...Can't we use equations to the upside motion(after it goes higher than A)..Is there something wrong in my first calculation..this problem is haert burning...thanks in advance..!

 

[shalikadm]

sorry for bump

 

[HallsofIvy]

I'm not sure what you mean by the "downside". The solution you give is for the weight moving both up and down. Take the derivative with respect to t, set it equal to 0 and solve for t. Those values of t will give the max and min values. The average value is half way between the max and min.

You could also do this by using trig identities to write $Ccos(\theta)+ Dsin(\theta)+ 2a$ as $A sin(\phi)+ 2a$ and the central point will be obvious.

 

[shalikadm]

I'm not sure what you mean by the "downside".

What I meant by upside and downside is motion above 'O' and motion below 'o'..
Note:the mass is projected upwards first.then it travels under gravity.then the string extends(pass 'A').so travels up in SHM..and down in SHM..then the string gets to its natural length...so engage in a free fall...then it passes 'o' under gravity...the string extends again after its natural length in the downward motion..so engage in SHM again..travels down in SHM...and up in SHM again..am I correct ?
When we find the center of SHM of a mass projected downwards,we get its equilibrium position (where T=mg) as the center...But when projected upward,I get something that doesn't true-center of SHM at 'O' where it's under a net force (gravity)..
What would u say about my calculation in the images above?Are they correct.?

 

[shalikadm]

solved​